If three Faraday of electricity is passed through the solutions of $A g {N O}_{3}, C u {S O}_{4}$ and $A u {C l}_{3}$ , the molar ratio of the metals deposited at the cathode will be:

1 : 1 : 1

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1 : 2 : 3

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3 : 2 : 1

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6 : 3 : 2

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Solution

The correct option is B $6 : 3 : 2$
$C u$ metal deposited as ${C u}^{2 +}$ ion
$A g$ metal deposited as ${A g}^{+}$ ion
$A u$ metal deposited as ${A u}^{3 +}$ ion
$∴$ to form 1 atom of $C u$ , $A g$ and $A u$
$↓$ $↓$ $↓$
${2 e}^{-}$ ${1 e}^{-}$ ${3 e}^{-}$
if 3 mol Faraday of electricity is passed,
then
1 mol of $A u$ , 3 mol of $A g$ and 3/2 mol of $C u$
i.e. $A g, C u, A u$
$3 : \frac{3}{2} : 1$
$6 : 3 : 2$

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